# How to determine the resistance in a electric heater

1. Sep 15, 2011

### ronaldinho52

hallo everyone,

I have a problem which i cant solve. Does somebody know how to solve this problem?? The question is:

A 220 V electric heater has two heating coils which can be switched such that either coil can be used independently or can be connected in series or parallel, yielding a total of four possible configurations.
The warmest setting corresponds to 2000 W and the coolest corresponds to 300 W.
R1 > R2

Determine the resistance of R1 [in Ω].

regards

2. Sep 15, 2011

### Hootenanny

Staff Emeritus
What have you tried so far? How do you think you can approach it?

3. Sep 15, 2011

### ronaldinho52

I've drawn four configurations;
1) two resistors parallel (R1 and R2) with a source voltage of 220 V
2) 1 resistor with a source voltage of 220 V
3) 1 resistor with a source voltage of 220 V
4) two resistors (R1 and R2) in series with a source voltage of 220 V

For each of these configurations I've computed the equivalent resistance and concluded that configuration 1 consumes the most power (so 2000 W) and config. 2 the least power (300 W). So for config. 1 I have 4 unknowns; R1, R2, I1, I2 and I have 3 equations; R1=U/I1 R2=U/I2 and finally P=(I1^2)*R1 + (I2^2)*R2. So I'm missing one equation to solve this problem and I do not know which one. I've also tried to use the last configuration and to relate these configs but I didn't get an answer that makes reason.

4. Sep 15, 2011

### Hootenanny

Staff Emeritus
You just need to think a little more. For example, in the parallel circuit you know that $V_1=V_2=V$. Thus,

$$P_p = V^2\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$$

And you already know the parallel circuit power consumption. Now, how about the single resistor circuit?

5. Sep 15, 2011

### Staff: Mentor

Wouldn't the least power occur when the net resistance is as large as possible? After all, P = V2/R.